hdu 4312 Meeting point-2(4级)
Meeting point-2
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 866 Accepted Submission(s): 472
Problem Description
It has been ten years since TJU-ACM established. And in this year all the retired TJU-ACMers want to get together to celebrate the tenth anniversary. Because the retired TJU-ACMers may live in different places around the world, it may be hard to find out where to celebrate this meeting in order to minimize the sum travel time of all the retired TJU-ACMers.
It's an opportunity to show yourself in front of your predecessors!
There is an infinite integer grid at which N retired TJU-ACMers have their houses on. They decide to unite at a common meeting place, which is someone's house. From any given cell, all 8 adjacent cells are reachable in 1 unit of time.
Eg: (x,y) can be reached from (x-1,y), (x+1,y), (x, y-1), (x, y+1), (x-1,y+1), (x-1,y-1), (x+1,y+1), (x+1,y-1).
Finding a common meeting place which minimizes the sum of the travel time of all the retired TJU-ACMers.
It's an opportunity to show yourself in front of your predecessors!
There is an infinite integer grid at which N retired TJU-ACMers have their houses on. They decide to unite at a common meeting place, which is someone's house. From any given cell, all 8 adjacent cells are reachable in 1 unit of time.
Eg: (x,y) can be reached from (x-1,y), (x+1,y), (x, y-1), (x, y+1), (x-1,y+1), (x-1,y-1), (x+1,y+1), (x+1,y-1).
Finding a common meeting place which minimizes the sum of the travel time of all the retired TJU-ACMers.
Input
The first line is an integer T represents there are T test cases. (0<T <=10)
For each test case, the first line is an integer n represents there are n retired TJU-ACMers. (0<n<=100000), the following n lines each contains two integers x, y coordinate of the i-th TJU-ACMer. (-10^9 <= x,y <= 10^9)
For each test case, the first line is an integer n represents there are n retired TJU-ACMers. (0<n<=100000), the following n lines each contains two integers x, y coordinate of the i-th TJU-ACMer. (-10^9 <= x,y <= 10^9)
Output
For each test case, output the minimal sum of travel times.
Sample Input
4 6 -4 -1 -1 -2 2 -4 0 2 0 3 5 -2 6 0 0 2 0 -5 -2 2 -2 -1 2 4 0 5 -5 1 -1 3 3 1 3 -1 1 -1 10 -1 -1 -3 2 -4 4 5 2 5 -4 3 -1 4 3 -1 -2 3 4 -2 2
Sample Output
20 15 14 38HintIn the first case, the meeting point is (0,2); the second is (0,0), the third is (1,-1) and the last is (-1,-1)
Author
TJU
Source
2012 Multi-University Training Contest 2
Recommend
zhuyuanchen520
思路：
max{|x|,|y|}=1/2(|x+y|+|x-y|);
因此将x变成x+y,y->x-y;就可以算曼哈顿距离了。
x,y到某点的曼哈顿距离可以分开算，因此，分开枚举以某点为中心的值球最小就好了。
预处理，kx[x]以某点 x为中心的所有x距离和，ky[]同理。kx[i]=kx[i-1]+(i-n+i)*(f[i].x-f[i-1].x);
就是将其按照 X，Y排序，答案在中间段取点，但应该枚举几个呢？？？这是个问题，所以有点狗屎
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
const int mm=2e5+9;
class node
{
public:
__int64 x,y;
int id;
} f[mm];
bool cmpx(node a,node b)
{
return a.x<b.x;
}
bool cmpy(node a,node b)
{
return a.y<b.y;
}
__int64 kx[mm],ky[mm];
__int64 sum;
__int64 aabs(__int64 x)
{
if(x<0)return -x;
return x;
}
int main()
{
int cas;__int64 n,xx,yy;
while(~scanf("%d",&cas))
{
while(cas--)
{
scanf("%I64d",&n);
for(int i=0; i<n; ++i)
{
scanf("%I64d%I64d",&xx,&yy);
f[i].x=xx-yy;f[i].y=xx+yy;
}
sort(f,f+n,cmpx);
kx[0]=0;
for(int i=0; i<n; ++i)
{
kx[0]+=aabs(f[i].x-f[0].x);
f[i].id=i;
}
for(__int64 i=1; i<n; ++i)
kx[i]=kx[i-1]+(i-n+i)*(f[i].x-f[i-1].x);
//for(int i=0;i<n;++i)
//cout<<" "<<kx[i]<<" ";puts("");
sort(f,f+n,cmpy);
ky[0]=0;
for(int i=0;i<n;++i)
ky[0]+=aabs(f[i].y-f[0].y);
for(__int64 i=1;i<n;++i)
ky[i]=ky[i-1]+(i+i-n)*(f[i].y-f[i-1].y);
//for(int i=0;i<n ;++i)
//cout<<" "<<ky[i]<<" ";puts("");
sum=6e18;
for(int i=0;i<n;++i)
{ //cout<<f[i].id<<" "<<i<<" "<<kx[f[i].id]<<" "<<ky[i]<<" "<<ky[i]+kx[f[i].id]<<endl;
if(sum>ky[i]+kx[f[i].id])sum=ky[i]+kx[f[i].id];
}
printf("%I64d\n",sum/2);
}
}
return 0;
}
///250以下基本就错了，以上基本就超了。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <set>
#include <algorithm>
using namespace std;
const int maxn = 1000010;
pair<int,int> p[maxn];
int abs1(int a) {
return a<0?-a:a;
}
int main()
{
int T,n;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
for(int i=0;i<n;i++)
scanf("%d %d",&p[i].first,&p[i].second);
sort(p,p+n);
int l = max(n/2-250,0);
int r = min(n/2+250,n);
__int64 ans = 999999999999999LL;
//cout << ans << endl;
for(int i=l;i<r;i++)
{
__int64 tmp = 0;
for(int j=0;j<n;j++)
tmp += max(abs1(p[j].first-p[i].first),abs1(p[j].second-p[i].second));
ans = min(ans,tmp);
}
printf("%I64d\n",ans);
}
return 0;
}
/*
3
10 88
0 1
0 5
1
0 2
0 0
1
1
1
*/